**problem**: Find the set of all values of \({m}\) such that \({\displaystyle {y} = {\frac{x^2-x}{1-mx}}}\) can take all real values.

**solution**: \({\displaystyle {y} = {\frac{x^2-x}{1-mx}}}\)

\({\Leftrightarrow}\) \({\displaystyle{y – myx = x^2 – x}}\)

\({\Leftrightarrow}\) \({x^2 + x(my – 1) – y = 0}\)

\({\Leftrightarrow}\) \({\displaystyle{x} = {\frac{1 – my {\pm} {\sqrt{(my-1)^2+4y}}}{2}}}\)

Now \({y}\) takes all real values if discriminant \({(my-1)^2 + 4y}\) is allways \({> 0}\).

So now we have to find the all values of \({m}\) such that \({(my-1)^2 + 4y}\) \({> 0}\) for all \({y}\) \({\in}\) |R.

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